Reverse process of XOR

Question

I was thinking, when two byte array is XOR'ed, is there any way to reverse the process? As I am curious to find the answer I found this code in the google which is the XOR code and this is the pseudocode of step1 (XOR code)

The XOR code:

public static byte[] XOR(byte[] first, byte[] second)
    {
        if (first.Length == second.Length)
        {
            byte[] result = new byte[first.Length];
            for (int i = 0; i < first.Length; i++)
            {
                result[i] = (byte)(first[i] ^ second[i]);
            }

            return result;
        }
        else
        {
            throw new ArgumentException();
        }
    }

The code works fine. Now down below is the pseudocode of step2 (reverse of XOR code)

I am very confused how to do this step2. How can I reverse the process to get back the original byte arrays? As the only input I have is the result. Is this ever possible? Is there any way to do this? If the answer is yes then how?

Answer 1

Forget about the arrays - just solve it for one bit.

Starting with the truth table for XOR:

a | b | c
0   0 | 0
0   1 | 1
1   0 | 1
1   1 | 0

If C (the output) is zero, can you figure out A and B? The answer is . . . no!

If you know that C is zero, then [A,B] was either [0,0] or [1,1]. The same thing happens if C is 1. You know that [A,B] was either [0,1] or [1,0].

Now, if you happen to know A and C, then you can figure out B. More generally, if you know two then you can figure out the 3rd. If you only know one then you can narrow down the other values to two choices, but that's as far as you can get without additional information

Answer 2

XOR is a reversible process; if you XOR the result with the same 2-byte array you used in the first example, you will get back the original. E.g. you have a two byte array Array1, and you XOR it with a two byte array Array2 to get the result Array3. If you want to get back Array1, simply take Array3 XOR Array2. Hope this helps